# I Differentiation of sin function where's my mistake?

1. Dec 21, 2017

### Hawksteinman

I was thinking and came up with this. I know it's wrong but can't find the mistake :(

dy/dx sin(x) = cos(x)
dy/dx sin(kx) = kcos(kx)

So dy/dx sin(3x) = 3cos(3x)
Now let Y = 3x
dy/dx sin(Y) = cos(Y) = cos(3x)

3cos(3x) = cos(3x)
3 = 1

Where is the mistake?

2. Dec 21, 2017

### Staff: Mentor

$\dfrac{d}{dx}\sin(Y) \neq \cos(Y)$
You've made the chain rule vanish by pretending $Y=y$.

3. Dec 21, 2017

### Staff: Mentor

Your notation doesn’t make sense.
dy/dx is the derivative of y with respect to x. “dy/dx sin(x)” is not a well-defined expression.

What you nean is d/dx sin(x). And suddenly the issue disappears:

d/dx sin(3x)=3cos(3x)
d/dx sin(y)=? - here we need the chain rule and d/dx 3x = 3:
d/dx sin(y) = cos(y) d/dx y = cos(y) * 3 = 3cos(3x)

4. Dec 21, 2017

### Hawksteinman

I haven't done the chain rule yet I'll need to look into that :)

5. Dec 21, 2017

### FactChecker

Basically:
If y =3x changes three times as fast as x, and sin(y) changes cos(y) times as fast as y, then sin(y)=sin(3x) changes 3*cos(y) = 3*cos(3x) times as fast as x.

6. Dec 21, 2017

### Staff: Mentor

As already noted, the above should be $\frac d {dx}\left(\sin(3x)\right) = 3\cos(3x)$
$\frac{dy}{dx}$ is the derivative of y with respect to x, so it is a thing, a noun.
$\frac d{dx}$ is the operator that signifies taking the derivative of something with respect to x. It is an action that hasn't completed yet, a verb. Don't confuse these two things.

Actually, you used the chain rule in the first line of what I quoted, above. The chain rule is what gives you that leading factor of 3.

7. Dec 21, 2017

### Hawksteinman

I don't know, I just used a table of derivatives :/

8. Dec 21, 2017

### Staff: Mentor

They were using the chain rule in the table.